
A body of mass $5\,kg$ moving with a velocity $10\,m{s^{ - 1}}$ collides with another body of the mass $20\,kg$ at rest and comes to rest. Velocity of the second body due to the collision is:-
A. $2.5\,m{s^{ - 1}}$
B. $5\,m{s^{ - 1}}$
C. $25\,m{s^{ - 1}}$
D. $2\,m{s^{ - 1}}$
Answer
164.7k+ views
Hint: We will use the principle of conservation of linear momentum to answer this problem, and then we will use equations using this principle to ascertain the velocity of the second body following the impact.
Formula used:
The principle of conservation of linear momentum says that Initial momentum of a system is always equal to final momentum of a system.
${P_i} = {P_f}$
where $P = mv$ denotes the momentum of a body defined as the product of mass of the body and the velocity of the body.
Complete step by step solution:
We have given that, the initial conditions of both bodies are ${m_1} = 5kg,{u_1} = 10\,m{s^{ - 1}}$ and ${m_2} = 20kg,{u_2} = 0$ so net initial momentum of the system is,
${P_i} = {m_1}{u_1} + {m_2}{u_2} \\
\Rightarrow {P_i} = 50\,kgm{s^{ - 1}} \\ $
And after the collision first body came to rest which means zero velocity therefore zero momentum and let velocity of second body is v then final momentum of the system is,
${P_f} = {m_2}v \\
\Rightarrow {P_f} = 20v \\ $
Now, using law of conservation of linear momentum we have,
${P_i} = {P_f} \\
\Rightarrow 50 = 20v \\
\therefore v = 2.5\,m{s^{ - 1}}$
Hence, the correct answer is option A.
Note: It should be remembered that, in addition to the law of conservation of momentum, there are two basic conservation laws: the law of conservation of energy, which states that energy is conserved, and the law of conservation of angular momentum in rotational dynamics, which states that the total angular momentum of the system is conserved when no external torque is applied to the body.
Formula used:
The principle of conservation of linear momentum says that Initial momentum of a system is always equal to final momentum of a system.
${P_i} = {P_f}$
where $P = mv$ denotes the momentum of a body defined as the product of mass of the body and the velocity of the body.
Complete step by step solution:
We have given that, the initial conditions of both bodies are ${m_1} = 5kg,{u_1} = 10\,m{s^{ - 1}}$ and ${m_2} = 20kg,{u_2} = 0$ so net initial momentum of the system is,
${P_i} = {m_1}{u_1} + {m_2}{u_2} \\
\Rightarrow {P_i} = 50\,kgm{s^{ - 1}} \\ $
And after the collision first body came to rest which means zero velocity therefore zero momentum and let velocity of second body is v then final momentum of the system is,
${P_f} = {m_2}v \\
\Rightarrow {P_f} = 20v \\ $
Now, using law of conservation of linear momentum we have,
${P_i} = {P_f} \\
\Rightarrow 50 = 20v \\
\therefore v = 2.5\,m{s^{ - 1}}$
Hence, the correct answer is option A.
Note: It should be remembered that, in addition to the law of conservation of momentum, there are two basic conservation laws: the law of conservation of energy, which states that energy is conserved, and the law of conservation of angular momentum in rotational dynamics, which states that the total angular momentum of the system is conserved when no external torque is applied to the body.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
